Find the range of the function f x = ln 8 - x 2 x - 2 .

Given a function f x = ln 8 - x 2 x - 2 A logarithmic function is defined when it's argument is always positive i.e. 8 - x 2 x - 2 > 0 Since, the square root is a positive quantity. ⇒ 8 - x 2 > 0 ⇒ x 2 - 8 < 0 ⇒ x + 2 2 x - 2 2 < 0 ⇒ x ∈ - 2 2 , 2 2 . . . . . . . . . i Also, numerator of L.H.S is positive and R.H.S is positive so, denominator of L.H.S should also be positive i.e. x - 2 > 0 ⇒ x > 2 . . . . . . . . . i i Both equations hold at a time so x ∈ 2 , 2 2 is the domain of f x . Since, 8 - x 2 x - 2 is continuous in the domain of f x as 8 - x 2 and x - 2 are also continuous in domain of f x . Finding limits at boundary points of domain : lim x → 2 + f x = lim x → 2 + ln 8 - x 2 x - 2 = lim h → 0 + ln 8 - 2 + h 2 2 + h - 2 = lim h → 0 + ln 8 - h 2 + 4 h + 4 h = lim h → 0 + ln 4 - 4 h - h 2 h = lim h → 0 + ln 2 h = + ∞ Also, lim x → 2 2 - f x = lim x → 2 2 - ln 8 - x 2 x - 2 = lim h → 0 + ln 8 - 2 2 - h 2 2 2 - h - 2 = lim h → 0 + ln 8 - h 2 - 4 2 h + 8 2 2 - 2 - h = lim h → 0 + ln 4 2 h - h 2 2 2 - 2 - h = - ∞ ∴ Range of f x is R .

Find the range of function f x = 1 16 - 4 x 2 - x .

f x = 1 16 - 4 x 2 - x Minimum value of x 2 - x occurs at x = - b 2 a = - - 1 2 = 1 2 . Minimum value of x 2 - x is 1 2 2 - 1 2 = - 1 4 . ⇒ x 2 - x ≥ - 1 4 ⇒ 4 x 2 - x ≥ 4 - 1 4 ⇒ 4 x 2 - x ≥ 1 2 ⇒ - 4 x 2 - x ≤ - 1 2 ⇒ 16 - 4 x 2 - x ≤ 16 - 1 2 ⇒ 16 - 4 x 2 - x ≤ 16 - 1 2 ⇒ 1 16 - 4 x 2 - x ≥ 1 16 - 1 2 ⇒ f x ≥ 1 16 - 1 2 f x ∈ 1 16 - 1 2 , ∞ .

Find the range of function f x = 1 2 - cos 3 x .

f x = 1 2 - cos 3 x As we know that range of cos x is - 1 , 1 . ⇒ - 1 ≤ cos x ≤ 1 ⇒ - 1 ≤ cos 3 x ≤ 1 ⇒ 1 ≥ - cos 3 x ≥ - 1 (Sign of inequality changes when multiplied by negative sign.) ⇒ 2 + 1 ≥ 2 - cos 3 x ≥ 2 - 1 ⇒ 3 ≥ 2 - cos 3 x ≥ 1 ⇒ 1 3 ≤ 1 2 - cos 3 x ≤ 1 ⇒ 1 3 ≤ f x ≤ 1 ⇒ f x ∈ 1 3 , 1 .

Find the range of function f x = x + 2 x 2 - 8 x - 4 .

f x = y = x + 2 x 2 - 8 x - 4 Cross multiply ⇒ y x 2 - 8 x - 4 = x + 2 ⇒ y x 2 - 8 x y - 4 y = x + 2 ⇒ y x 2 - 8 x y - x - 4 y - 2 = 0 ⇒ y x 2 - 8 y + 1 x - 4 y + 2 = 0 Here, x is real So, the equation will have real roots ⇒ D ≥ 0 ⇒ - 8 y + 1 2 - 4 y - 4 y + 2 ≥ 0 ⇒ 64 y 2 + 16 y + 1 + 16 y 2 + 8 y ≥ 0 ⇒ 80 y 2 + 24 y + 1 ≥ 0 ⇒ 80 y 2 + 20 y + 4 y + 1 ≥ 0 ⇒ 4 y + 1 20 y + 1 ≥ 0 ∴ y ∈ - ∞ , - 1 4 ∪ - 1 20 , ∞ Cross-check where co-efficient of x 2 = 0 ⇒ y = 0 . It is possible when x = - 2 . So, y ∈ - ∞ , - 1 4 ∪ - 1 20 , ∞ .

Find the range of the function f x = x 2 - 2 x + 4 x 2 + 2 x + 4 .

f x = y = x 2 - 2 x + 4 x 2 + 2 x + 4 Cross multiplying, we get, y x 2 + 2 x y + 4 y = x 2 - 2 x + 4 ⇒ y - 1 x 2 + 2 y + 2 x + 4 y - 4 = 0 x is real ⇒ Above equation have real roots ⇒ D ≥ 0 ⇒ 2 y + 2 2 - 4 y - 1 4 y - 4 ≥ 0 ⇒ 4 y 2 + 8 y + 4 - 16 y 2 + 32 y - 16 ≥ 0 ⇒ y 2 + 2 y + 1 - 4 y 2 + 8 y - 4 ≥ 0 ⇒ - 3 y 2 + 10 y - 3 ≥ 0 ⇒ 3 y 2 - 10 y + 3 ≤ 0 ⇒ 3 y - 1 y - 3 ≤ 0 ⇒ y ∈ 1 3 , 3 Cross-check where co-efficient of x 2 = 0 ⇒ y - 1 = 0 ⇒ y = 1 . It is possible when x = 0 . So, y ∈ 1 3 , 3 .

Find the range of function f x = 3 sin π 2 16 - x 2 .

f x = 3 sin π 2 16 - x 2 As, x 2 ≥ 0 ⇒ - x 2 ≤ 0 ⇒ π 2 16 - x 2 ≤ π 2 16 ⇒ 0 ≤ π 2 16 - x 2 ≤ π 2 16 ⇒ 0 ≤ π 2 16 - x 2 ≤ π 4 sin 0 ≤ sin π 2 16 - x 2 ≤ sin π 4 ⇒ 0 ≤ sin π 2 16 - x 2 ≤ 1 2 0 ≤ 3 sin π 2 16 - x 2 ≤ 3 2 ⇒ 0 ≤ f x ≤ 3 2 f x ∈ 0 , 3 2 .